I need to understand a shell code which uses the following command to fetch directions from a source to destination using GOOGLE MAPS API:
wget --no-parent -O - https://maps.googleapis.com/maps/api/directions/json?origin=$begin\&destination=$finish\&sensor=false > new.txt
Next we fetch the following line of the output:
**"html_instructions" : "Head \u003cb\u003enorthwest\u003c/b\u003e"**
grep -n html_instructions new.txt > new1.txt
Can somebody please tell me the meaning of using:
sed -e 's/\\u003cb//g'
etc in the following command:
sed -e 's/\\u003cb//g' -e 's/\\u003e//g' -e 's/\\u003c\/b//g' -e 's/\\u003c//g' -e 's/div.*div//g' -e 's/.*://g' -e 's/"//g' -e 's/ "//g' new1.txt > new2.txt
Which outputs Head northwest only.
Thanks in advance!
sed -e 's/\\u003cb//g' -e 's/\\u003e//g' -e 's/\\u003c\/b//g' -e 's/\\u003c//g' -e 's/div.*div//g' -e 's/.*://g' -e 's/"//g' -e 's/ "//g' new1.txt > new2.txt
The string after each -e is a sed command. The sed command s/\\u003cb//g searches for all occurrences of the unicode character 003CB (which is a greek small letter upsilon with dialytika) and replaces it with nothing. In other words, it remove the character from the string.
Thus, that sed command removes every occurrence of unicode characters 003cb, u003e, and u003c from the lines and new1.txt and sends the output to new2.txt.
Additionally, s/div.*div//g causes any string that begins and ends with "div" to be removed. The command s/.*://g removes any text from the beginning of the line to the last colon in the line. s/"//g removes the every occurence of the double-quote character. s/ "//g removes every occurrence of space followed by double-quote.
In general, the sed command s/new/old/ searches for the first occurrence of new and replaces it with old. With a g appended at the end, as in s/new/old/g, it makes the substitution globally: looks for every occurrence of new and replaces it with old. Adding a lot of power to these commands, new may be a regular expression. Consider s/.*://g. The dot character has the special meaning of "any character at all". The star character means zero or more of the preceding character. Thus the regular expression.*:` means zero or more of any characters followed by a colon.
You can take all in one go with awk:
awk -F\" '/html_instructions/ {gsub(/(\\u003(c|cb|e)|\/b)/,x);print $4}'
Head northwest
So whole line should be:
wget --no-parent -O - https://maps.googleapis.com/maps/api/directions/json?origin=$begin\&destination=$finish\&sensor=false | awk -F\" '/html_instructions/ {gsub(/(\\u003(c|cb|e)|\/b)/,x);print $4}'
Head northwest
to get it into a variable
d=$(wget --no-parent -O - https://maps.googleapis.com/maps/api/directions/json?origin=$begin\&destination=$finish\&sensor=false | awk -F\" '/html_instructions/ {gsub(/(\\u003(c|cb|e)|\/b)/,x);print $4}')
echo $d
Head northwest
Related
Given a file, for example:
potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789
I'd like to grep for all lines that start with potato: but only pipe the numbers that follow potato:. So in the above example, the output would be:
1234
5432
How can I do that?
grep 'potato:' file.txt | sed 's/^.*: //'
grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).
or
grep 'potato:' file.txt | cut -d\ -f2
For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).
or
grep 'potato:' file.txt | awk '{print $2}'
For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.
or
grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'
All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.
or
awk '{if(/potato:/) print $2}' < file.txt
The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.
or
perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt
The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).
Or use regex assertions: grep -oP '(?<=potato: ).*' file.txt
grep -Po 'potato:\s\K.*' file
-P to use Perl regular expression
-o to output only the match
\s to match the space after potato:
\K to omit the match
.* to match rest of the string(s)
sed -n 's/^potato:[[:space:]]*//p' file.txt
One can think of Grep as a restricted Sed, or of Sed as a generalized Grep. In this case, Sed is one good, lightweight tool that does what you want -- though, of course, there exist several other reasonable ways to do it, too.
This will print everything after each match, on that same line only:
perl -lne 'print $1 if /^potato:\s*(.*)/' file.txt
This will do the same, except it will also print all subsequent lines:
perl -lne 'if ($found){print} elsif (/^potato:\s*(.*)/){print $1; $found++}' file.txt
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code
You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.
Try this (semicolons are there to allow you to put it all on one line):
$ while read line;
do
if [[ "${line%%:\ *}" == "potato" ]];
then
echo ${line##*:\ };
fi;
done< file.txt
## tells bash to delete the longest match of ": " in $line from the front.
$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789
or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.
$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi
The substring to split on is ":\ " because the space character must be escaped with the backslash.
You can find more like these at the linux documentation project.
Modern BASH has support for regular expressions:
while read -r line; do
if [[ $line =~ ^potato:\ ([0-9]+) ]]; then
echo "${BASH_REMATCH[1]}"
fi
done
grep potato file | grep -o "[0-9].*"
Given a file, for example:
potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789
I'd like to grep for all lines that start with potato: but only pipe the numbers that follow potato:. So in the above example, the output would be:
1234
5432
How can I do that?
grep 'potato:' file.txt | sed 's/^.*: //'
grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).
or
grep 'potato:' file.txt | cut -d\ -f2
For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).
or
grep 'potato:' file.txt | awk '{print $2}'
For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.
or
grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'
All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.
or
awk '{if(/potato:/) print $2}' < file.txt
The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.
or
perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt
The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).
Or use regex assertions: grep -oP '(?<=potato: ).*' file.txt
grep -Po 'potato:\s\K.*' file
-P to use Perl regular expression
-o to output only the match
\s to match the space after potato:
\K to omit the match
.* to match rest of the string(s)
sed -n 's/^potato:[[:space:]]*//p' file.txt
One can think of Grep as a restricted Sed, or of Sed as a generalized Grep. In this case, Sed is one good, lightweight tool that does what you want -- though, of course, there exist several other reasonable ways to do it, too.
This will print everything after each match, on that same line only:
perl -lne 'print $1 if /^potato:\s*(.*)/' file.txt
This will do the same, except it will also print all subsequent lines:
perl -lne 'if ($found){print} elsif (/^potato:\s*(.*)/){print $1; $found++}' file.txt
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code
You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.
Try this (semicolons are there to allow you to put it all on one line):
$ while read line;
do
if [[ "${line%%:\ *}" == "potato" ]];
then
echo ${line##*:\ };
fi;
done< file.txt
## tells bash to delete the longest match of ": " in $line from the front.
$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789
or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.
$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi
The substring to split on is ":\ " because the space character must be escaped with the backslash.
You can find more like these at the linux documentation project.
Modern BASH has support for regular expressions:
while read -r line; do
if [[ $line =~ ^potato:\ ([0-9]+) ]]; then
echo "${BASH_REMATCH[1]}"
fi
done
grep potato file | grep -o "[0-9].*"
I want to delete three words with a special character on a line such as
Input:
\cf4 \cb6 1749,1789 \cb3 \
Output:
1749,1789
I have tried a couple sed and grep statements but so far none have worked, mainly due to the character \.
My unsuccessful attempt:
sed -i 's/ [.\c ] //g' inputfile.ext >output file.ext
Awk accepts a regex Field Separator (in this case, comma or space):
$ awk -F'[ ,]' '$0 = $3 "." $4' <<< '\cf4 \cb6 1749,1789 \cb3 \'
1749.1789
-F'[ ,]' - Use a single character from the set space/comma as Field Separator
$0 = $3 "." $4 - If we can set the entire line $0 to Field 3 $4 followed by a literal period "." followed by Field 4 $4, do the default behavior (print entire line)
Replace <<< 'input' with file if every line of that file has the same delimeters (spaces/comma) and number of fields. If your input file is more complex than the sample you shared, please edit your question to show actual input.
The backslash is a special meta-character that confuses bash.
We treat it like any other meta-character, by escaping it, with--you guessed it--a backslash!
But first, we need to grep this pattern out of our file
grep '\\... \\... [0-9]+,[0-9]+ \\... \\' our_file # Close enough!
Now, just sed out those pesky backslashes
| sed -e 's/\\//g' # Don't forget the g, otherwise it'll only strip out 1 backlash
Now, finally, sed out the clusters of 2 alpha followed by a number and a space!
| sed -e 's/[a-z][a-z][0-9] //g'
And, finally....
grep '\\... \\... [0-9]+,[0-9]+ \\... \\' our_file | sed -e 's/\\//g' | sed -e 's/[a-z][a-z][0-9] //g'
Output:
1749,1789
My guess is you are having trouble because you have backslashes in input and can't figure out how to get backslashes into your regex. Since backslashes are escape characters to shell and regex you end up having to type four backslashes to get one into your regex.
Ben Van Camp already posted an answer that uses single quotes to make the escaping a little easier; however I shall now post an answer that simply avoids the problem altogether.
grep -o '[0-9]*,[0-9]*' | tr , .
Locks on to the comma and selects the digits on either side and outputs the number. Alternately if comma is not guaranteed we can do it this way:
egrep -o ' [0-9,]*|^[0-9,]*' | tr , . | tr -d ' '
Both of these assume there's only one usable number per line.
$ awk '{sub(/,/,".",$3); print $3}' file
1749.1789
$ sed 's/\([^ ]* \)\{2\}\([^ ]*\).*/\2/; s/,/./' file
1749.1789
I have a text file that's about 300KB in size. I want to remove all lines from this file that begin with the letter "P". This is what I've been using:
> cat file.txt | egrep -v P*
That isn't outputting to console. I can use cat on the file without another other commands and it prints out fine. My final intention being to:
> cat file.txt | egrep -v P* > new.txt
No error appears, it just doesn't print anything out and if I run the 2nd command, new.txt is empty.
I should say I'm running Windows 7 with Cygwin installed.
Explanation
use ^ to anchor your pattern to the beginning of the line ;
delete lines matching the pattern using sed and the d flag.
Solution #1
cat file.txt | sed '/^P/d'
Better solution
Use sed-only:
sed '/^P/d' file.txt > new.txt
With awk:
awk '!/^P/' file.txt
Explanation
The condition starts with an ! (negation), that negates the following pattern ;
/^P/ means "match all lines starting with a capital P",
So, the pattern is negated to "ignore lines starting with a capital P".
Finally, it leverage awk's behavior when { … } (action block) is missing, that is to print the record validating the condition.
So, to rephrase, it ignores lines starting with a capital P and print everything else.
Note
sed is line oriented and awk column oriented. For your case you should use the first one, see Edouard Lopez's reponse.
Use sed with inplace substitution (for GNU sed, will also for your cygwin)
sed -i '/^P/d' file.txt
BSD (Mac) sed
sed -i '' '/^P/d' file.txt
Use start of line mark and quotes:
cat file.txt | egrep -v '^P.*'
P* means P zero or more times so together with -v gives you no lines
^P.* means start of line, then P, and any char zero or more times
Quoting is needed to prevent shell expansion.
This can be shortened to
egrep -v ^P file.txt
because .* is not needed, therefore quoting is not needed and egrep can read data from file.
As we don't use extended regular expressions grep will also work fine
grep -v ^P file.txt
Finally
grep -v ^P file.txt > new.txt
This works:
cat file.txt | egrep -v -e '^P'
-e indicates expression.
This bash script is supposed to remove leading whitespace from grep results:
#!/bin/bash
grep --color=always $# | sed -r -e's/:[[:space:]]*/:/'
But it doesn't match the whitespace. If I change the substitution text to "-", that shows up in the output, but it still never removes the whitespace. I've tried it without the "*", escaping the "*", with "+", etc, but nothing works. Does anyone know why not?
(I'm using sed version 4.2.1 on Ubuntu 12.04.)
Thanks all, this is my modified script, which shows grep color and also trims leading blanks:
#!/bin/bash
grep --color=always $# | sed -r -e's/[[:space:]]+//'
You need to remove the --color option for this to work. The color codes confuse sed:
grep $# | sed -r -e's/:[[:space:]]*/:/'
The color information output by grep takes the form of special character sequences (see answers to this StackOverflow question), so if the colon is colored and the whitespace isn't, or vice versa, then that means that one of these character sequences will be between them, so sed will not see them as adjacent characters.
The character class \s will match the whitespace characters and
For example:
$ sed -e "s/\s\{3,\}/ /g" inputFile
will substitute every sequence of at least 3 whitespaces with two spaces.
grep --color=always $# |sed 's/^ //g'
Removes leading white spaces.